Chapter 5, Problem 136AP

Interpretation Introduction

Interpretation:

The final temperature of $\text{Ca}{\left(\text{OH}\right)}_2$ and the enthalpy of formation of calcium hydroxide are to be calculated.

Concept Introduction:

The standard enthalpy change for a reaction is the amount of enthalpy change thatoccurs at the standard conditions.

The standard enthalpy of the reaction is to be determined using the equation as given below:

$\Delta H{°}_{rxn}={\displaystyle \sum n}\Delta H_f^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_f^{°}\left(\text{reactants}\right).$

Here, the stoichiometric coefficients are represented by m for reactants and n for products and theenthalpies of formation at the standard conditions are represented by $\Delta H_f^{°}$.

The standard enthalpy of formation is the amount of heat change when one mole of compound is formed from its integral elements that are present in their standard states.

The heat is related to the change in temperature of a material, which is as follows:

$q=sm\Delta T.$

Here, q represents heat, s signifies specific heat, m shows mass, and $\Delta T$ is the change in the temperature of the material.

Answer to Problem 136AP

Solution:

a)

$758\text{°C}$

b)

$-986.6\text{ kJ}/\text{mol}$

Explanation of Solution

a)The final temperature of the product $\text{Ca}{\left(\text{OH}\right)}_2$.

Mass of water is $m_{\text{water}}=500\text{ g}\text{.}$

The reaction for quicklime is as follows:

${\text{CaCO}}_3\left(s\right)\to \text{CaO}\left(s\right)+{\text{CO}}_2\left(g\right)\Delta H°=177.8\text{ kJ}/\text{mol}\text{.}$

The reaction for slaked lime is as follows:

$\text{CaO}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\Delta H°=-65.2\text{ kJ}/\text{mol}\text{.}$

The molar mass of ${\text{H}}_2\text{O}\left(l\right)$ is $M_{H_{2}O}=18\text{ g}/\text{mol}\text{.}$

The number of moles is calculated by the expression, which isas follows:

$n=\frac{m}{M}$. …… (1)

Substitute $500\text{ g}$ for $m$ and $18\text{ g}/\text{mol}$ for $M_{H_{2}O}$ in the equation (1).

$\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{500\cancel{\text{g}}}{18\cancel{\text{g}}/\text{mol}}\\ =27.75\text{ mol}\text{.}\end{array}$

As water and $\text{CaO}$ are equimolar, so

$\begin{array}{c}{n}_{\text{CaO}}=n_{{\text{H}}_2\text{O}}\\ =27.75\text{ mol}\end{array}$

Similarly, the mole ratio of $\text{Ca}{\left(\text{OH}\right)}_2$ and ${\text{H}}_2\text{O}\left(l\right)$ is $1:1$. So,

$\begin{array}{c}{n}_{\text{Ca}{\left(\text{OH}\right)}_2}=n_{{\text{H}}_2\text{O}}\\ =27.75\text{ mol}\text{.}\end{array}$

The molar mass of $\text{Ca}{\left(\text{OH}\right)}_2$ is $M_{\text{Ca}{\left(\text{OH}\right)}_2}=74\text{ g}/\text{mol}\text{.}$

To calculate the mass of $\text{Ca}{\left(\text{OH}\right)}_2$ the equation (1) can be written as follows:

$m_{\text{Ca}{\left(\text{OH}\right)}_2}=n_{\text{Ca}{\left(\text{OH}\right)}_2}\times M_{\text{Ca}{\left(\text{OH}\right)}_2}.$

Substitute $27.75\text{ mol}$ for $n_{\text{Ca}{\left(\text{OH}\right)}_2}$ and $74\text{ g}/\text{mol}$ for $M_{\text{Ca}{\left(\text{OH}\right)}_2}$ in the above equation.

$\begin{array}{c}{m}_{\text{Ca}{\left(\text{OH}\right)}_2}=\left(27.75\cancel{\text{mol}}\right)\times \left(74\text{ g}/\cancel{\text{mol}}\right)\\ =2053.5\text{ g}\\ \approx \text{2056 g}\\ \approx 2.056\times {10}^3\text{ g}\text{.}\end{array}$

So, the heat produced by $500\text{ g}$ of water is calculated by the expression, which is as follows:

$q=n\times \Delta H.$

Substitute $27.75\text{ mol}$ for $n$ and $-65.2\text{ kJ}/\text{mol}$ for $\Delta H$ in the above equation.

$\begin{array}{c}q=\left(27.75\cancel{\text{mol}}\right)\times \left(-65.2\text{ kJ}/\cancel{\text{mol}}\right)\\ =-1809.3\text{ kJ}\\ =-1.809\times {10}^3\text{ kJ}\\ =-1.809\times {10}^6\text{ J}\text{.}\end{array}$

The relation between theheat and change in the temperature of a materialis given as follows:

$q=sm\Delta T.$

The specific heat(s) of $\text{Ca}{\left(\text{OH}\right)}_2$ is $1.20\text{ J}/\text{g}\text{.}°\text{C}$.

Substitute $1.809\times {10}^6\text{ J}$ for $q$, $2.056\times {10}^3\text{ g}$ for $m,$ and $1.20\text{ J}/\text{g}\text{.°C}$ for $s$ in the above equation.

$\begin{array}{c}1.809\times {10}^6\cancel{\text{J}}=\left(1.20\cancel{\text{J}}/\cancel{\text{g}\text{.}}\text{°C}\right)\times \left(2.056\times {10}^3\cancel{\text{g}}\right)\times \Delta T\\ =2.4672\times {10}^3\times \Delta T.\end{array}$

The temperature change is calculated as follows:

$\begin{array}{c}\Delta T=\frac{1.809\times {10}^6}{2.4672\times {10}^3}\text{ °C}\\ =733\text{°C}\text{.}\end{array}$

The change in the temperature of the material is given by the expression, which is as follows:

$\begin{array}{c}\Delta T=T_{final}-T_{initial};\\ T_{final}=\Delta T+T_{initial}.\end{array}$

So, the final temperature is calculated as follows:

$\begin{array}{c}{T}_{final}=733\text{°C}+25\text{°C}\\ =758\text{°C}\text{.}\end{array}$

b) The standard enthalpy of formation of $\text{Ca}{\left(\text{OH}\right)}_2.$

Standard enthalpy of formation of $\left[\text{CaO}\left(s\right)\right]$ and $\left[{\text{H}}_2\text{O}\left(l\right)\right]$ are $-635.6\text{ kJ}/\text{mol}$ and $-285.8\text{ kJ}/\text{mol,}$ respectively.

The reaction for slaked lime is as follows:

$\text{CaO}\left(s\right)+{\text{H}}_2\text{O}\left(l\right)\to \text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\Delta H°=-65.2\text{ kJ}/\text{mol,}$

From appendix 2, the enthalpies of formation values of the given compounds are given as follows:

$\begin{array}{l}\Delta H{°}_f\left[\text{CaO}\left(s\right)\right]=-635.6\text{ kJ}/\text{mol;}\\ \Delta H{°}_f\left[{\text{H}}_2\text{O}\left(l\right)\right]=-285.8\text{ kJ}/\text{mol}\text{.}\end{array}$

The standard enthalpy of reaction is calculated by the expression, as follows:

$\Delta H{°}_{rxn}={\displaystyle \sum n}\Delta H_f^{°}\left(\text{products}\right)-{\displaystyle \sum m}\Delta H_f^{°}\left(\text{reactants}\right).$

Now, the standard enthalpy of the given reaction is calculated as follows:

$\begin{array}{c}\Delta H{°}_{rxn}=\left\{\Delta H{°}_f\left[\text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\right]\right\}-\left\{\Delta H{°}_f\left[\text{CaO}\left(s\right)\right]+\Delta H{°}_f\left[{\text{H}}_2\text{O}\left(l\right) \right]\right\}\\ -65.2\text{ kJ}/\text{mol}=\Delta H{°}_f\left[\text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\right]-\left[-635.6\text{ kJ}/\text{mol}+\left(-285.8\text{ kJ}/\text{mol}\right)\right]\\ =\Delta H{°}_f\left[\text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\right]-\left[-921.4\text{ kJ}/\text{mol}\right].\end{array}$

This can be rearranged and solved as follows:

$\begin{array}{c}\Delta H{°}_f\left[\text{Ca}{\left(\text{OH}\right)}_2\left(s\right)\right]=-65.2\text{ kJ}/\text{mol}-921.4\text{ kJ}/\text{mol}\\ =-986.6\text{ kJ}/\text{mol}\text{.}\end{array}$